Consider the following reaction.

On estimation of bromine in 1.00 g of R using Carius method, the amount of AgBr formed (in g) is ______.

[Given : Atomic mass of H = 1, C = 12, O = 16, P = 31, Br = 80, Ag = 108]

1. Identify the Major Product R

When 4-bromobenzyl alcohol is treated with red phosphorus and bromine (red P / Br₂), it acts as a brominating agent (PBr₃ formed in situ). This reagent selectively replaces the aliphatic hydroxyl group (-OH) with a bromine atom (-Br). The aromatic ring-bound bromine remains unaffected.

• Starting material: 4-bromobenzyl alcohol (C₇H₇OBr)
• Major product R: 4-bromobenzyl bromide
• Molecular formula of R: C₇H₆Br₂

2. Calculate the Molar Mass of Compound R (C₇H₆Br₂)

Using the given atomic masses:

• Carbon (C): 7 atoms × 12 = 84
• Hydrogen (H): 6 atoms × 1 = 6
• Bromine (Br): 2 atoms × 80 = 160

Molar mass of R = 84 + 6 + 160 = 250 g/mol

3. Analysis of Carius Method for Estimation of Bromine

In the Carius method, all bromine atoms present in the organic compound are converted quantitatively into Silver Bromide (AgBr) precipitates.

1 mole of C₇H₆Br₂ contains 2 moles of Br atoms, which will yield 2 moles of AgBr.

• Molar mass of AgBr = Mass of Ag + Mass of Br = 108 + 80 = 188 g/mol

4. Stoichiometric Calculation

From the relationship established above:

• 1 mole of R (250 g) gives 2 moles of AgBr (2 × 188 g = 376 g)

Now, we calculate the amount of AgBr formed from 1.00 g of R:

• Mass of AgBr formed = (376 g / 250 g) × 1.00 g
• Mass of AgBr formed = 1.504 g

Rounding off to two decimal places as shown in standard numerical formats:

• Mass of AgBr formed = 1.50 g